A man standing on a turn-table is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is 75% of that with outstret

Question

A man standing on a turn-table is rotating at a certain angular frequency with his arms outstretched. He suddenly folds his arms. If his moment of inertia with folded arms is 75% of that with outstretched arms, then his rotational kinetic energy will

TestHub · Accepted Answer

By the law of conservation of angular momentum, $L_{1} = L_{2} = L$ .

The rotational kinetic energy is given by $K = \frac{L ^{2}}{2 I}$ .

Let the initial moment of inertia be $I_{1} = I$ .

The final moment of inertia is $I_{2} = 75% of I = \frac{3}{4} I$ .

Initial kinetic energy, $K_{1} = \frac{L ^{2}}{2 I}$ .

Final kinetic energy, $K_{2} = \frac{L ^{2}}{2 I _{2}} = \frac{L ^{2}}{2 ( \frac{3}{4} I )} = \frac{4}{3} (\frac{L ^{2}}{2 I}) = \frac{4}{3} K_{1}$ .

Percentage increase in kinetic energy $= \frac{K _{2} - K _{1}}{K _{1}} \times 100% = \frac{\frac{4}{3} K _{1} - K _{1}}{K _{1}} \times 100% = \frac{1}{3} \times 100% = 33.3%$ .

Answer: increase by 33.3 %

Physics - Rotation Question with Solution | TestHub

Options:

Doubts & Discussion